Determine which of the following are open sets in $\mathbb{R}l$. In each case, prove your assertion

Question

I have very little knowledge of set theory and proof writing. This is a problem from “Introduction to Topology: Pure and Applied” by Colin Adams and Robert Franzosa.

DEFINITION

On $\mathbb{R}$, let $\beta=\{[a,b)\subset\mathbb{R}\;|\;a<b\}$. The collection $B$ is a basis for a topology on $\mathbb{R}$. We call the topology generated by this basis the lower limit topology since each basis element contains its lower limit. We denote $\mathbb{R}$ with this topology by $\mathbb{R}l$.

EXAMPLE

The intervals $[0,2)$ and $(0,2)$ are both open in $\mathbb{R}l$. The former is open since it is a basis element; the latter is open since it is the union of the basis elements $B_i=[1/i, 2)$, where $i=1,2,3\dots$

PROBLEM

Determine which of the following are open sets in $\mathbb{R}l$. In each case, prove your assertion.

$A=[4,5)$

$B=\{3\}$

$C=[1,2]$

$D=(7,8)$

THOUGHTS

We can see from the example that $A$ and $D$ are open. For $B$, I can't think of a way that $\{3\}$ could be the union of any of the basis elements, but I don't know how I would prove that. For $C$, I don't know how we could arrive at something of the form $[a,b]$ from the union of intervals in the form $[a,b)$. As always, I appreciate any help.

Answer 1

You are correct in asserting that $\{3\}$ is not open. Let $U$ be an open set containing $3$. Then, because $U$ is open in $\mathbb{R}$, and $B$ generates the lower limit topology, it follows (by the definition of a topology generated by a basis), that there is a basis element $B_0$ containing $3$ that is contained in $U$. That is, $3 \in B_0 \subset U$. But, since $B_0$ is a basis element, it is of the form $[a,b)$. In other words, $3 \in [a,b) \iff a \leq 3 < b$. But since there is a real number $c$ between $3$ and $b$, we conclude that $c \in [a,b) \subset U$. So, we have found another element of $U$ that is not $3$. In other words, any open set containing $3$ must also contain another real number, so the singleton $\{3\}$ cannot be open.

For $[1,2]$, we note $2 \in [1,2]$, and if it was open, there should be a basis element $[a,b)$ containing $2$ such that $2 \in [a,b) \subset [1,2]$, but clearly this is impossible, since if $2 \in [a,b)$, then $a \leq 2 < b$, but since $[a,b) \subset [1,2]$, we have $b \leq 2$, so we have the contradiction that $2 < b \leq 2$.

As a side note: I think you may have gotten tunnel-vision in this problem. Yes, seeing elements of a topology as a union of basis elements is often helpful and will take you a long way! However, never forget about your basic definitions :)

Answer 2

Indeed $A$ is (basic) open by definition, and $(7,8)$ is open in the same way as the example: $(7,8) = \bigcup_{7< x < 8} [x,8)$, a union of basic open subsets.

All basic open sets are uncountable, so all open subsets are uncountable hence $\{3\}$ is not open.

Suppose $[1,2]$ were a union of basic open sets. One of these must contain $2$, so there is some $[a,b)$ basic open with $2 \in [a,b)$ and $[a,b) \subseteq [1,2]$ (the latter, as it's one of the set whose union is $[1,2]$ so in particular, they're all subsets of $[1,2]$) But then $b > a \ge 2$ and so for any $c$ with $a < c < b$ we have that $c \in [a,b)$ but $c \notin [1,2]$ contradiction. Basically: basic open subsets that contain $2$ always contain points to the right of $2$ and so $2$ is not an interior point of $[1,2]$.

Answer 3

For $B$: if $\{3\}$ were a union of the basis elements, it would contain some interval of the form $[a,b); a<b$. But clearly no such interval is a subset of $\{3\}$ (for example, all such intervals have more than one element whereas $\{3\}$ has only one).

For $C$: Let $\mathcal{F}$ be the family of intervals whose union is $C$ (so $\cup\mathcal{F}=C$). Then $\exists I\in\mathcal{F}:2\in I$. But since $I$ is of the form $[a,b)$ and $2<b$, there must be some $x>2$ in $I$ as well. So this $x$ will be in $C$, a contradiction.